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How do I show that a divisible group $G\neq 0$ is not free?

I know that divisible means that for all elements $g$ in an abelian group $G$ and $n\in\mathbb{N}$ there is an element $a\in G$ such that $na=g$. But I don't know what to do with this.

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Any nontrivial free abelian group admits a surjective morphism onto $\mathbb Z$ (map one basis element to $1$ and do not even care about what the other basis elements do).

A divisible group admits only the zero homomorphism to $\mathbb Z$, since the homomorphic image of a divisible group is again divisible. The only divisible subgroup of $\mathbb Z$ is the trivial subgroup.

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